MATH 151 Midterm: MATH 151 TAMU Y2014 2014c X2A Solutions

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turquoisegnu128

31 Jan 2019

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Fri, 31/oct c(cid:13)2014 art belmonte f (x) f (a: (b) two limit de nitions of f (a) are lim x a f (a + h) f (a) and lim h 0. For f (x) = (2x + 1)3/2 and a = 1 , these are represented by choices (iii) and (ii), respectively. The other is rubbish. h x a: (c) for x 6= 0, f (x) = 5x4 + 1, (b) now v (t) = s (t) = (cid:0)t 2 + 1(cid:1) (1) t (2t) 5 . (t 2 + 1)2: (d) the so-called normal line y = x+3. So the slope of the tangent line at p (1, f (1)) is f (1) = 1/m = 2. With g (x) = x2 f (x), we have g (x) = 2x f (x) + x2 f (x). Since p is on the normal line, f (1) = 1+3.

Related Questions

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